Potential energy solutions, Summaries of Physics

Potential energy solutions . Minimum potential energy solutions

Typology: Summaries

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Problem 1
The spring is un-stretched when θ = 30o. At any
position of the pendulum, the spring remains
horizontal. If the spring constant is k = 50 N/m, at
what position will the system be in equilibrium.
(Soln: No need to solve this problem. It involves
Newton-Raphson like method for actual
solution)
Monday 19 April 2010
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Problem 1

• The spring is un-stretched when θ = 30

o

. At any

position of the pendulum, the spring remains

horizontal. If the spring constant is k = 50 N/m, at

what position will the system be in equilibrium.

( Soln: No need to solve this problem. It involves

Newton-Raphson like method for actual

solution)

Problem 2

• If the springs are un-

stretched when θ = θ

o

find the angle θ when

the weight W is applied

on the system. Use the

method of minimum

potential energy.

θ = 0 ; cos(θ) = cos(θ

W

(k

+ k

)a

Problem 3

• Two bars are attached to a single spring of

constant k that is un-stretched when the bars are

vertical. Determine the range of values of P for

which the equilibrium of the system is stable in the

position shown.

P ≤

ka

P ≤

ka

4 L

( stability both for parts (a) and (b))

Problem 4

• The horizontal bar AD is attached to two springs of constant

k and is in equilibrium in the position shown. Determine the

range of values of the magnitude P of the two equal and

opposite horizontal forces P and –P for which the equilibrium

position is stable if (a) AB = CD, (b) if AB = 2CD.

o

, stable; θ = 207

o

unstable

Problem 6

• Determine the equilibrium values of θ and the

stability of equilibrium at each point for the

unbalanced wheel on the 10

o

incline. Static friction

is sufficient to prevent slipping. The mass center is

at G.

Extra Problem 1

• The uniform disk of radius R and mass m rolls without

slipping on the fixed cylinder surface of radius 2R. Fastened

to the disk is a lead cylinder also of mass m with its center

located a distance b from the center O of the disk. Determine

the minimum value of b for which the disk will remain in

stable equilibrium on the cylindrical surface.

b

min

R

Extra Problem 3

  • One of the critical requirements in the design of an artificial leg for

an amputee is to prevent the knee joint from buckling under load

when the leg is straight. As a first approximation, simulate the

artificial leg by the two light links with a torsion spring at their

common joint. The spring develops a torque M = K β, which Is

proportional to the angle of the bend β at the joint. Determine the

minimum value of K which will ensure the stability of the knee joint

for β = 0.

K

min

mgl

Extra Problem 4

• The horizontal bar BEH is pinned to collar E and to

vertical bars AC and GI. The collar can slide freely

on bar DF. Determine the range of values of Q for

which the equilibrium of the system is stable in the

position shown when a = 480mm, b = 400mm, and

P = 600N.

Q > 432 N

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,

reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited

distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,

you are using it without permission.

1798

PROBLEM 10.

A slender rod AB , of weight W , is attached to two blocks A

and B that can move freely in the guides shown. Knowing that

the spring is unstretched when y =0,determine the value of

y corresponding to equilibrium when W =80 N, l =500 mm,

and k =600 N/m.

SOLUTION

Deflection of spring = s , where

2 2

2 2

= + −

=

s l y l

ds y

dy l y

Potential energy:

( )

2

2 2

2 2

2 2

y

V ks W

dV ds

ks W

dy dy

dV y

k l y l W

dy l y

l

k y W

l y

Equilibrium

2 2

= ¨^ − ¸ =

dV l W

y

dy k l y

Now W = 80 N, l = 0.500 m, and k =600 N/m

Then

2 2

0.500 m 1 (80 N)

2 (600 N/m) (0.500)

y

y

or

2

y

y

Solving numerically, y =0.357 m y =357 mmW